Indefinite integral of $\int (g(x))^{3}(f(x))^{3}(g'(x)f(x) + g(x)f'(x))\ dx$ I want to find the integral $\int [g(x)]^{3}[f(x)]^{3}(g'(x)f(x) + g(x)f'(x))\ dx$
Here is what I have.
\begin{align}
  \int [g(x)]^{3}[f(x)]^{3}(g'(x)f(x) + g(x)f'(x))\ dx
= \int ([g(x)]^{3}[f(x)]^{3}g'(x)f(x) + [g(x)]^{3}[f(x)]^{3}g(x)f'(x))\ dx
= \int ([g(x)]^{3}[f(x)]^{4}g'(x) + [g(x)]^{4}[f(x)]^{3}f'(x)) = ?
 \end{align}
The answer seems to be $\frac{1}{4}[g(x)]^{4}[f(x)]^{4}$. However, I am not exactly sure how to get the answer.
 A: For brevity, I'll write $f=f(x)$, and $g=g(x)$, so $f^3$ is actually $[f(x)]^3$ and $f'$ is actually $f'(x)$ (same for $g$).
To start, break up the integral into a sum of two integrals. Then apply integration by parts using $u'=f^3f'$ and $v=g^4$, so that $u=\frac14f^4$ and $g'=4g^3g'$. You'll find that the integral that results from integration by parts, and the second integral in the original sum cancel out, as follows:
$$
\begin{align}
\int{f^3g^3(f'g + fg')}&=\int{f^3g^4f'}+\int{f^4g^3g'} \\
& = \frac14f^4g^4-\frac44\int f^4g^3g'+\int{f^4g^3g'} \\
& = \frac14f^4g^4+\left(\int{f^4g^3g'}-\int{f^4g^3g'}\right) \\
& = \frac14f^4g^4
\end{align}
$$
Another way of solving this integral is to apply integration by parts using $u'=f'g + fg'$ and $v=f^3g^3$ as follows:
$$
\begin{align}
\int f^3g^3(f'g + fg') & = f^4g^4 - 3\int fg(3f^2f'g^3 + 3f^3g^2g') \\
& = f^4g^4 - 3\int f^3g^3(f'g + fg')
\end{align}
$$
Notice that the integral on the RHS is the same as the original one. Thus, rearranging yields
$$
4\int f^3g^3(f'g + fg') = f^4g^4 \\
\int f^3g^3(f'g + fg') = \frac14 f^4g^4
$$
There are a lot of nice symmetries to exploit in this problem.
A: First note that we can write $f^3g^3=(fg)^3$.  
Then, recall the product rule for differentiation $(fg)'=fg'+f'g$.  
Therefore, we have
$$f^3g^3(fg'+f'g)=(fg)^3(fg)'$$
Now, let $u=fg$.  Then, 
$$\begin{align}
\int f^3g^3(fg'+f'g)\,dx&=\int (fg)^3(fg)'\,dx\\\\
&=\int u^3u'\,dx\\\\
&=\int u^3 du\\\\
&=\frac14u^4+C\\\\
&=\frac14 f^4g^4+C
\end{align}$$
